Permutation polynomials, fractional polynomials, and algebraic curves

arXiv:1708.04841v1 [math.CO] 16 Aug 2017

PERMUTATION POLYNOMIALS, FRACTIONAL POLYNOMIALS, AND ALGEBRAIC CURVES DANIELE BARTOLI 1,† AND MASSIMO GIULIETTI 2

Abstract. In this note we prove a conjecture by Li, Qu, Li, and Fu on permutation trinomials over F2k 3 . In addition, new examples and generalizations of some families of permutation polynomials of F3k and F5k are given. We also study permutation quadrinomials of type Axq(q−1)+1 + Bx2(q−1)+1 + Cxq + x. Our method is based on the investigation of an algebraic curve associated with a fractional polynomial over a finite field.

1. Introduction Let q = ph be a prime power. A polynomial f (x) ∈ Fq [x] is a permutation polynomial (PP) if it is a bijection of the finite field Fq into itself. On the other hand, each permutation of Fq can be expressed as a polynomial over Fq . Permutation polynomials were first studied by Hermite and Dickson; see [3, 6]. In general it is not difficult to construct a random PP for a given field Fq . Particular, simple structures or additional extraordinary properties are usually required by applications of PPs in other areas of mathematics and engineering, such as cryptography, coding theory, or combinatorial designs. Permutation polynomials meeting these criteria are usually difficult to find. For a deeper introduction on the connections of PPs with other fields of mathematics we refer to [12, 7] and the references therein. In this work we deal with a particular class of PP. For a prime p and a positive integer m, let Fpm be the finite field with pm elements. Given a polynomial h(x) over Fpm , a divisor d of pm − 1, and an integer r with 1 ≤ r < (pm − 1)/d, let pm −1 . fr,d,h (x) = xr h x d A useful criterion to decide weather fr,d,h permutes Fpm is the following.

Theorem 1.1. [13, 17] The polynomial fr,d,h (x) is a PP of Fpm if and only if m gcd(r, (pm − 1)/d) = 1 and xr h(x)(p −1)/d permutes the set µd of the d-th roots of unity in Fpm . Pℓ i Let q = pn . For h(x) = i=0 ai x a polynomial over Fq2 , by Theorem 1.1 xr h xq−1 permutes Fq2 if and only if xr h (x)q−1 permutes µq+1 . If this is the case and in addition h(x) ∈ Fq [x], then for each z ∈ µq+1 we have that q−1

z r h (z)

q

= zr

e h(z) h(1/z) (h(z)) = zr = z r−ℓ , h(z) h(z) h(z)

Date: August 17, 2017. Key words and phrases. Permutation polynomials; fractional permutation polynomials. 1

DANIELE BARTOLI 1,† AND MASSIMO GIULIETTI 2

2

Pℓ e where deg(h) = ℓ and e h(x) = i=0 aℓ−i xi . We call the rational function xr−ℓ h(x) h(x) the fractional polynomial associated with the PP xr h xq−1 . Conversely, given a e h(x) fractional polynomial xr−ℓ h(x) which permutes µq+1 we call xr h xq−1 the associated permutation polynomial. A standard approach to the problem of deciding whether a polynomial f (x) is a PP is the investigation of the plane algebraic curve f (x) − f (y) Cf : = 0; x−y in fact, f is a PP over Fpm if and only if Cf has no Fpm -rational point (a, b) with a 6= b. In the case where pm = q 2 and f is of type fr,q+1,h with h ∈ Fq [x], it can be more effective to study the curve, with degree lower than Cf , defined by the equation e e h(y) h(x) − y r−ℓ =0 h(x) h(y) and check whether it has some Fq2 -rational points (a, b) with a 6= b and aq+1 = bq+1 = 1. In this note we use both such methods to solve some conjectures on permutation polynomials, shorten several proofs that have recently appeared in the literature, and provide some new examples of permutation quadrinomials and permutation functions of type α(x)/β(x), where α, β ∈ Fq [x]. xr−ℓ

2. On some conjectures on permutation polynomials In [9] the authors presented the following conjecture about permutation trinomials in characteristic 3. Conjecture 2.1. [9, Conjecture 5.1] (1) Let q = 3k , k even, and f (x) = xℓq+ℓ+5 + x(ℓ+5)q+ℓ − x(ℓ−1)q+ℓ+6 , where gcd(5 + 2ℓ, q − 1) = 1. Then f (x) is a permutation trinomial over Fq 2 . (2) Let q = 3k , f (x) = xℓq+ℓ+1 − x(ℓ+4)q+ℓ−3 − x(ℓ−2)q+ℓ+3 and gcd(1 + 2ℓ, q − 1) = 1. Then f (x) is a permutation trinomial over Fq2 . (3) Let q = 3k , f (x) = xℓq+ℓ+1 + x(ℓ+2)q+ℓ−1 − x(ℓ−2)q+ℓ+3 and gcd(1 + 2ℓ, q − 1) = 1. Then f (x) is a permutation trinomial over Fq2 if k 6≡ 2 (mod 4). It has been noticed in [8] that by Theorem 1.1 the above conjecture can be rephrased as follows. Conjecture 2.2. [9, Conjecture 5.2] −x7 +x6 +x x6 +x−1 . Then g(x) permutes x6 +x4 −1 −x7 +x3 +x . Then g(x) permutes µq+1 .

(1) Let q = 3k , k even and g(x) = k

(2) Let q = 3 and g(x) =

µq+1 .

PERMUTATION POLYNOMIALS, FRACTIONAL POLYNOMIALS, AND ALGEBRAIC CURVES3

(3) Let q = 3k and g(x) = (mod 4).

−x5 +x3 +x x4 +x2 −1 .

Then g(x) permutes µq+1 if k 6≡ 2

In [8] the author settles Conjecture 2.2(2) and Conjecture 2.2(3) by determining some quadratic factors of a five-degree polynomial and a seven-degree polynomial. The aim of this section is twofold. On yhe one hand, we settle Conjecture 2.2(1) using different arguments with the respect to those of [8]. On the other hand, we show that basic tools from Algebraic Geometry can be very useful when dealing with permutation polynomials (see also [2, 1]); in particular they can provide shorter and less technical proofs. Theorem 2.3. Let q = 3k , k even. The function g(x) =

−x7 +x6 +x x6 +x−1

permutes µq+1 .

Proof. Note that if x ∈ µq+1 , then q q 6 7 x +x−1 x6 − x5 − 1 −1 −x6 − x + 1 −x + x6 + x = , = −x = x6 + x − 1 x6 + x − 1 x −x6 + x5 + 1 −x7 + x6 + x that is g(µq+1 ) ⊂ µq+1 . We have only to show that g(x) is injective on µq+1 . Let C be the plane curve given by the affine equation H(x, y) = 0, where H(x, y) = (−x7 + x6 + x)(y 6 + y − 1) − (x6 + x − 1)(−y 7 + y 6 + y) = 0. The function g(x) permutes µq+1 if and only if the curve C does not have any point (x, y) ∈ µ2q+1 off the line x = y. The polynomial H factorizes as H(x, y) = (x − y)F (x, y)G(x, y), where F (x, y) = x3 y 3 + ω 2 x3 y 2 + ω 5 x3 y + ω 5 x3 + ω 6 x2 y 3 + x2 y 2 + 2x2 y + ω 5 x2 +ω 7 xy 3 + 2xy 2 + xy + ω 2 x + ω 7 y 3 + ω 7 y 2 + ω 6 y + 1 and G(x, y) = x3 y 3 + ω 6 x3 y 2 + ω 7 x3 y + ω 7 x3 + ω 2 x2 y 3 + x2 y 2 + 2x2 y + ω 7 x2 +ω 5 xy 3 + 2xy 2 + xy + ω 6 x + ω 5 y 3 + ω 5 y 2 + ω 2 y + 1, for some primitive element ω of F9 . Note that both F (x, y) and G(x, y) are polynomials defined over F3k , since k is even. Suppose that x, y ∈ µq+1 are such that F (x, y) = 0. Then 0 = x3 y3 F (x, y)q = 3 3 x y F (1/x, 1/y) = x3 y3 + ω 6 x3 y2 + ω 7 x3 y + ω 7 x3 + ω 2 x2 y3 + x2 y2 + 2x2 y + ω 7 x2 +ω 5 xy 3 + 2xy 2 + xy + ω 6 x + ω 5 y 3 + ω 5 y 2 + ω 2 y + 1. The resultant between F (x, y) and x3 y 3 F (1/x, 1/y) with the respect to y is (x+ω 2 )9 (x+ω 6 )9 . This implies that the common points of the curves with equations F (x, y) = 0 and x3 y 3 F (1/x, 1/y) = 0 belong to the lines x = 2ω 2 , 2ω 6 , but in this case xq = x 6= 1/x. Therefore no points (x, y) ∈ µ2q+1 satisfy F (x, y) = 0. A similar 7 +x6 +x argument holds for G(x, y). Then the function g(x) = −x x6 +x−1 permutes µq+1 . Remark 2.4. Similar arguments can be used to settle the Conjectures 2.2(2) and 2.2(3). In the former case, the curve splits into six conic components, apart from x − y = 0, defined by xy + η 2 x + η 15 y + 1 = 0,

xy + η 5 x + η 18 y + 1 = 0,

xy + η 6 x + η 19 y + 1 = 0


4

xy + η 15 x + η 2 y + 1 = 0, xy + η 18 x + η 5 y + 1 = 0, xy + η 19 x + η 6 y + 1 = 0, where η is some primitive element of F27 . In the latter case the curve splits into four conic components defined over F81 , namely xy + ξ 25 x + ξ 65 y + 1 = 0, xy + ξ 35 x + ξ 75 y + 1 = 0, xy + ξ 65 x + ξ 25 y + 1 = 0, xy + ξ 75 x + ξ 35 y + 1 = 0; here ξ is a primitive element of F81 . In both cases it can be easily proved that no points (x, y) ∈ µ2q+1 with x 6= y belong to a component of the curve, and therefore the corresponding fractional functions g(x) permute µq+1 . We now provide alternative and shorter proofs of two results from [11], referring to two conjectures presented in [14]. 2

2

(x −2) Proposition 2.5. Let q = 5k , k even. Then g(x) = −x (x 2 +2)2 permutes µq+1 .

Proof. It is enough to prove that g(x) is injective over µq+1 . The corresponding plane algebrai curve is −x(x2 − 2)2 (y 2 + 2)2 + y(y 2 − 2)2 (x2 + 2)2 = 0. x−y The equation of C can be rewritten as C:

F1 (x, y) · F2 (x, y) = 0 where F1 (x, y) = x2 y 2 + ω 3 x2 y + 3x2 + ω 3 xy 2 + xy + ω 15 x + 3y 2 + ω 15 y + 1 and F2 (x, y) = x2 y 2 + ω 15 x2 y + 3x2 + ω 15 xy 2 + xy + ω 3 x + 3y 2 + ω 3 y + 1, for a primitive element ω of F25 . A point (x, y) ∈ µ2q+1 belongs to a component Fi (x, y) = 0 if and only if satisfies both Fi (x, y) = Fi (1/x, 1/y) = 0. Now the resultant of the two polynomials Fi (x, y) and x2 y 2 Fi (1/x, 1/y) with the respect to y is always R(x) = 4x8 + 4x6 + 3x4 + 4x2 + 4. The polynomial R(x) factorizes as Q j j∈J (x + ω ), where J = {3, 4, 8, 9, 15, 16, 20, 21}. None of its roots is in µq+1 and therefore C does not contains points of type (x, y) ∈ µ2q+1 ; that is, g(x) permutes µq+1 . 2

2

−x+2) Proposition 2.6. Let q = 5k , k odd. Then g(x) = x (x (x2 +x+2)2 permutes Fq .

Proof. It is enough to show that g(x) is injective over Fq . The corresponding curve is x(x2 − x + 2)2 (y 2 + y + 2)2 − y(y 2 − y + 2)2 (x2 + x + 2)2 C : = 0. x−y This curve as an (affine) Fq -rational point (x, y), x 6= y, if and only if g(x) is a permutation of Fq . The equation of C can be written as F1 (x, y)F2 (x, y) = 0, where F1 (x, y) = (x2 y 2 + ω 7 x2 y + ω 2 x2 + ω 7 xy 2 + ω 3 xy + ωx + ω 2 y 2 + ωy + 4) F2 (x, y) = (x2 y 2 + ω 11 x2 y + ω 10 x2 + ω 11 xy 2 + ω 15 xy + ω 5 x + ω 10 y 2 + ω 5 y + 4),


for some primitive element ω of F25 . Note that the curves Fi (x, y) = 0 are not defined over Fq , since k is odd. The unique Fq -rational points of C satisfy F1 (x, y) = F2 (x, y) = 0. The resultant between F1 (x, y) and F2 (x, y) with the respect to y is R(x) = 4x8 + x7 + 4x6 + x5 + 3x4 + 3x3 + x2 + 2x + 4 and its roots are ω i , i ∈ {1, 4, 5, 13, 14, 17, 20, 22}. None of them is in F5k since k is odd and the claim follows. Using the same argument one can easily prove [10, Theorem 3.2] and [10, Theorem 3.25.]. Namely, when q = 2k the fractional polynomials x8 + x7 + x5 + x3 + x x8 + x7 + x6 + x5 + x3 + x2 + x , x7 + x6 + x5 + x3 + x2 + x + 1 x7 + x5 + x3 + x + 1 7 permute µq+1 . Let F27 = hwi, where w + w + 1 = 0. The equations of the associated curves can be written as 6 Y i i (xy + w19·2 x + w19·2 y + 1) = 0 i=0

and

6 Y

i

i

(xy + w3·2 x + w3·2 y + 1) = 0,

i=0 ξ+w respectively. Each component can be written as y = ξx+1 , for some ξ ∈ F27 . It n q is easily seen that if q = 2 with n ≡ 0 (mod 7) then y = 1/y for any x ∈ µq+1 , since ξ ∈ Fq . On the other hand if n 6≡ 0 (mod 7) then ξ ∈ / Fq and y q = 1/y if only if y = x = 1. Therefore both the fractional polynomials permute µq+1 if and only if q = 2n with n 6≡ 0 (mod 7). We end this section by discussin briefly two conjectures about permutation trinomials presented by Gupta and Sharma [5] and proved in [16, 15]: m m • the polynomial f (x) = x5 + x3·2 +2 + x4·2 +1 ∈ F22m [x] is a permutation trinomial over F22m if and only if m ≡ 2 (mod 4); m m • the polynomial f (x) = x5 + x2 +4 + x5·2 ∈ F22m [x] is a permutation trinomial over F22m if and only if m ≡ 2 (mod 4). The proofs given in [16, 15] rely on the fact that the rational functions

x + x2 + x5 1 + x + x3 1 + x + x5 , , 1 + x3 + x4 1 + x2 + x3 1 + x4 + x5 or their inverses permute µ2m +1 . This follows quite easily from considering the 1+x+x5 associated curves. For instance, the curve associated with the function 1+x 4 +x5 is (x5 + x4 + x)(y 4 + y + 1) + (y 5 + y 4 + y)(x4 + x + 1) = 0, x+y which splits as (xy + ωx + ω 4 y + 1)(xy + ω 2 x + ω 8 y + 1)(xy + ω 4 x + ωy + 1)(xy + ω 8 x + ω 2 y + 1) = 0, for some primitive element ω of F16 ; this means that the components of the curve are not defined over F2m if m ≡ 2 (mod 4). Whence, it is easy to show that no pairs (x, y) ∈ µ2m +1 , x 6= y belong to such a curve.


6

2

2

(x +Ax+B) Table 1. Permutation polynomials of type x (x 2 +Cx+D)2 of F5k , k odd, A, B, C, D ∈ F5

[4, 2, 1, 2] [4, 2, 3, 3] [2, 3, 4, 2] [2, 3, 3, 3] [1, 2, 4, 2] [1, 2, 2, 3] [3, 3, 1, 2] [3, 3, 2, 3] [0, 3, 0, 2] [0, 2, 0, 3]

Roots ω i of R(x)

F1 (x, y) and F2 (x, y)

A, B, C, D x2 y 2

ω 7 x2 y

ω 2 x2

ω 7 xy 2

ω 3 xy

ω2 y2

+ + + + + ωx + + ωy + 4, i = 1, 2, 5, 8, 10, 13, 16, 17 x2 y 2 + ω 11 x2 y + ω 10 x2 + ω 11 xy 2 + ω 15 xy + ω 5 x + ω 10 y 2 + ω 5 y + 4 x2 y 2 + ωx2 y + ω 14 x2 + ωxy 2 + ω 16 xy + ω 13 x + ω 14 y 2 + ω 13 y + 1, i = 1, 4, 5, 7, 8, 11, 16, 20 x2 y 2 + ω 5 x2 y + ω 22 x2 + ω 5 xy 2 + ω 8 xy + ω 17 x + ω 22 y 2 + ω 17 y + 1 x2 y 2 + ω 19 x2 y + ω 2 x2 + ω 19 xy 2 + ω 4 xy + ω 19 x + ω 2 y 2 + ω 19 y + 1, i = 1, 2, 5, 10, 14, 19, 22, 23 x2 y 2 + ω 23 x2 y + ω 10 x2 + ω 23 xy 2 + ω 20 xy + ω 23 x + ω 10 y 2 + ω 23 y + 1 x2 y 2 + ωx2 y + ω 14 x2 + ωxy 2 + ω 15 xy + ω 7 x + ω 14 y 2 + ω 7 y + 4, i = 2, 4, 7, 10, 11, 19, 20, 23 x2 y 2 + ω 5 x2 y + ω 22 x2 + ω 5 xy 2 + ω 3 xy + ω 11 x + ω 22 y 2 + ω 11 y + 4 2 2 19 2 2 2 19 2 3 13 2 2 13 x y + ω x y + ω x + ω xy + ω xy + ω x + ω y + ω y + 4, i = 1, 4, 5, 13, 14, 17, 20, 22 x2 y 2 + ω 23 x2 y + ω 10 x2 + ω 23 xy 2 + ω 15 xy + ω 17 x + ω 10 y 2 + ω 17 y + 4 2 2 13 2 14 2 13 2 16 14 2 x y + ω x y + ω x + ω xy + ω xy + ωx + ω y + ωy + 1, i = 4, 8, 13, 16, 17, 19, 20, 23 x2 y 2 + ω 17 x2 y + ω 22 x2 + ω 17 xy 2 + ω 8 xy + ω 5 x + ω 22 y 2 + ω 5 y + 1 2 2 7 2 2 2 7 2 4 7 2 2 7 x y + ω x y + ω x + ω xy + ω xy + ω x + ω y + ω y + 1, i = 2, 7, 10, 11, 13, 14, 17, 22 x2 y 2 + ω 11 x2 y + ω 10 x2 + ω 11 xy 2 + ω 20 xy + ω 11 x + ω 10 y 2 + ω 11 y + 1 2 2 13 2 14 2 13 2 15 19 14 2 19 x y + ω x y + ω x + ω xy + ω xy + ω x + ω y + ω y + 4, i = 7, 8, 11, 14, 16, 19, 22, 23 x2 y 2 + ω 17 x2 y + ω 22 x2 + ω 17 xy 2 + ω 3 xy + ω 23 x + ω 22 y 2 + ω 23 y + 4 2 2 3 2 2 3 2 15 2 15 x y + ω x y + 3x + ω xy + xy + ω x + 3y + ω y + 1, i = 3, 4, 8, 9, 15, 16, 20, 21 x2 y 2 + ω 15 x2 y + 3x2 + ω 15 xy 2 + xy + ω 3 x + 3y 2 + ω 3 y + 1 2 2 9 2 2 9 2 9 2 9 x y + ω x y + 2x + ω xy + 4xy + ω x + 2y + ω y + 1, i = 2, 3, 9, 10, 14, 15, 21, 22 x2 y 2 + ω 21 x2 y + 2x2 + ω 21 xy 2 + 4xy + ω 21 x + 2y 2 + ω 21 y + 1

3. New examples of fractional permutations In this section we extend some results of Section 2 to other fractional polynomials. We begin with the case p = 5. Proposition 3.1. Let q = 5k , k odd, and A, B, C, D ∈ F5 . Then gA,B,C,D (x) = (x2 +Ax+B)2 x (x 2 +Cx+D)2 permutes Fq in all the cases listed in Table 1. Proof. In all the cases listed in Table 1 ω stands for a primitive element of F25 . The curve corresponding to the fractional polynomial has two components F1 (x, y) = 0 and F2 (x, y) = 0 not defined over Fqk , but over Fq2k . One can argue as in the proof of Proposition 2.5. The last column of the table indicates the roots of the resultant R(x) of F1 (x, y) and F2 (x, y) with the respect to y. It is easily seen that in all these cases such roots do not belong to Fqk since k is odd; therefore, the curve associated with gA,B,C,D does not have any Fqk -rational points with distinct coordinates. Now we turn our attention to the case F3k . We consider functions of type gA1 ,A2 ,A3 ,A4 ,A5 ,A6 (x) =

−x7 + A1 x6 + A2 x5 + A3 x4 + A4 x3 + A5 x2 + A6 x) , (A6 x6 + A5 x5 + A4 x4 + A3 x3 + A2 x2 + A1 x − 1)

for Ai ∈ F3 and A1 6= 0. In Table 2 ω stands for a primitive element of F9 . The components of the associated curve over F9 are indicated. Note that every quadratic component in the table is absolutely irreducible. It is easily seen that their only F3 -rational points lie also on the line x = y. and therefore if k is odd these conics do not have rational points off the line x = y. It is worth noting that all the conics for k odd have points in µ23k +1 and therefore the corresponding g(x) does not permute µ3k +1 .


Table 2. Factorization over F32 of the curve associated with the function gA1 ,A2 ,A3 ,A4 ,A5 ,A6 (x) A1 , A2 , A3 , A4 , A5 , A6

[1, 1, 0, 2, 1, 1]

[2, 2, 1, 1, 2, 1]

[2, 2, 1, 2, 0, 1]

Factors A1 , A2 , A3 , A4 , A5 , A6 3 y + ω, y + ω y + ω5, y + ω7 x + ω, x + ω 3 [1, 2, 2, 2, 0, 1] x + ω5, x + ω7 xy + ωx + ω 3 y + 1 xy + ω 3 x + ωy + 1 y + ω2, y + ω6 x + ω2, x + ω6 [1, 2, 2, 1, 1, 1] xy + ω 5 x + ω 7 y + 1 7 5 xy + ω x + ω y + 1 y2 + ω5y + 1 y2 + ω7y + 1 xy + ωx + ω 3 y + 1 [2, 1, 0, 2, 2, 1] xy + ω 3 x + ωy + 1 2 5 x +ω x+1 x2 + ω 7 x + 1

Factors y + ωy + 1 y2 + ω3y + 1 xy + ω 5 x + ω 7 y + 1 xy + ω 7 x + ω 5 y + 1 x2 + ωx + 1 x2 + ω 3 x + 1 y + ω2, y + ω6 x + ω2, x + ω6 xy + ωx + ω 3 y + 1 xy + ω 3 x + ωy + 1 y + ω, y + ω 3 y + ω5, y + ω7 x + ω, x + ω 3 x + ω5, x + ω7 xy + ω 5 x + ω 7 y + 1 xy + ω 7 x + ω 5 y + 1 2

e

Finally, we deal with fractional functions of type hh , for odd characteristic p and deg(h) = 3. In particular, we are able to provide conditions on the coefficients of h e which ensures that hh is a permutation of µq+1 . This could be useful to investigate the permutation property of the associated polynomials. Proposition 3.2. Let p > 2. Let A, B, C ∈ Fq , A, B 6= 0. Let hA,B,C (x) = e h

(x)

A,B,C Bx3 + Cx2 + x + A The function fA,B,C (x) = hA,B,C (x) permutes µq+1 in the following cases: • A2 − AC − B 2 + B = 0, and AT 2 + (1 − B)T + A ∈ Fq [T ] has distinct roots in Fq ; • q ≡ 1 (mod 3), B = (3AC + C 2 − 1)/3, and the polynomial 3T 2 − 3(3A + C)T + (3A + C)2 − 1 ∈ Fq [T ] has distinct roots in Fq .

Proof.

• If p is odd and A2 − AC − B 2 + B = 0 then the equation of the associated curve CfA,B,C can be written as (Ax2 + (1 − B)x + A)(Ay 2 + (1 − B)y + A) = 0.

Let k1 and k2 the distinct roots of AT 2 + (1 − B)T + A; then the curve splits into the four lines x = k1 , x = k2 , y = k1 , and y = k2 . Since k1 , k2 ∈ Fq , these lines do not have points in µ2q+1 . In fact, Fq ∩ µq+1 = {±1} and k1 , k2 = ±1 would imply k1 = k2 . • Let p 6= 3 and assume that B = (3AC + C 2 − 1)/3. Then the equation of the associated curve CfA,B,C can be written as (xy + αx + (3A + C − α)y + 1)(xy + (3A + C − α)x + αy + 1) = 0 where α satisfies 3α2 − 3(3A + C)α + (3A + C)2 − 1 = 0.


8

By assumption α ∈ Fq . The conic xy + αx + (3A + C − α)y + 1 = 0 has a point (x0 , y0 ) in µ2q+1 if and only if q α + x0 αx0 + 1 q =− y0 = − x0 + 3A + C − α 1 + (3A + C − α)x0 equals 1/y0 =

x0 +3A+C−α . αx0 +1

Then

(α + x0 )(αx0 + 1) − ((3A + C − α) + x0 )((3A + C − α)x0 + 1) = 0 and hence (3A + C − 2α)(x20 + (3A + C)x0 + 1) = 0 for some x0 ∈ µq+1 . Since the roots of F (T ) = 3T 2 − 3(3A + C)T + (3A + C)2 − 1 are distinct, 3A + C − 2α = 0 cannot hold. Also, the roots of T 2 + (3A + C)T + 1 = 0 are in Fq since q ≡ 1 (mod 3) ensures that −3 is a square in Fq (see e.g. [4, Lemma 4.5]); hence, x0 ∈ {±1}. This implies 3A + C = ±2 and F (T ) = 3(T ± 1)2 , a contradiction. Corollary 3.3. Let p > 2, A, B, C ∈ Fq , A, B 6= 0. The polynomial FA,B,C (x) = Axq(q−1)+1 + Bx2(q−1)+1 + Cxq−1+1 + x permutes Fq2 if either (i) A2 − AC − B 2 + B = 0 and AT 2 + (1 − B)T + A has distinct roots in Fq , or (ii) q ≡ 1 (mod 3), B = (3AC + C 2 − 1)/3, and the polynomial F (T ) = 3T 2 − 3(3A + C)T + (3A + C)2 − 1 has distinct roots in Fq . Proof. By Theorem 1.1 FA,B,C (x) permutes Fq2 if and only if x(Axq + Bx2 + Cx + 3 +x2 +Cx+B 1)q−1 permutes µq+1 . On µq+1 the previous function equals Ax Bx3 +Cx2 +x+A . By Proposition 3.2 the assertion follows. We now show that the second condition of Corollary 3.3(i) can be replaced by ∈ / Fq .

B−1 A

Proposition 3.4. Let A, B, C ∈ Fq2 , A, B 6= 0. If A2 − AC − B 2 + B = 0 and B−1 / Fq then the polynomial FA,B,C (x) = Axq(q−1)+1 + Bx2(q−1)+1 + Cxq−1+1 + x A ∈ permutes Fq2 . Proof. Let fA,B,C be the fractional polynomial associated with the polynomial FA,B,C . We know by Proposition 3.2 that in this case the equation of the corresponding curve can be written as (Ax2 + (1 − B)x + A)(Ay 2 + (1 − B)y + A) = 0. Note that the roots of AT 2 + (1 − B)T + A are of type {z, 1/z} and therefore they belong to µq+1 if and only if (B − 1)/A = −z − z −1 = −z − z q for some z ∈ µq+1 . Note that z + z q ∈ Fq and by assumption this is not possible. 4. Acknowledgment The authors were supported in part by Ministry for Education, University and Research of Italy (MIUR) (Project PRIN 2012 Geometrie di Galois e strutture di incidenza”) and by the Italian National Group for Algebraic and Geometric Structures and their Applications (GNSAGA - INdAM).


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